\newproblem{lay:5_1_20}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.1.20}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Without calculation, find one eigenvalue and two linearly independent vectors of $A=\begin{pmatrix}2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2\end{pmatrix}$. Justify your answer.
}{
  % Solution
	The determinant of $A$ is zero because it has duplicated rows. On the other hand, the determinant is the product of the matrix eigenvalues, so at least one of
	the eigenvalues of $A$ must be zero. The eigenspace associated to the eigenvalue $\lambda=0$ is the set of vectors satisfying $A\mathbf{x}=\mathbf{0}$. Since the three rows
	are the same, we can eliminate the last two by subtracting the first one. To find the eigenvectors we note that if we subtract the first column to the second column we get a null
	vector (the corresponding eigenvector is $(-1,1,0)$). Similarly, if we subtract the first column to the third column, we again get a null vector (a second eigenvector is $(-1,0,1)$).
}
\useproblem{lay:5_1_20}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
